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Proof step 14 from proof get-sublist (previous step)

n→i(m) < j, 0 < j, ∀ j0, r0, i0, st0. ¬ j0 < n→i(m) → sublist(0, m, getarray(r0, i0, j0, st0)) = getarray(r0, i0, n→i(m), st0) ⊦ sublist(0, m + 1, getarray(r, i, j, st)) = getarray(r, i, n→i(m) + 1, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ vals ≠ [] → sublist(0, n + 1, vals) = vals .first + sublist(0, n, vals .rest)
Proof step 3:
⊦ getarray(r, i, j, st) = [] ↔ j ≤ 0
Proof step 13:
getarray(r, i, j, st) ≠ [], n→i(m) < j, 0 < j, ∀ j0, r0, i0, st0. ¬ j0 < n→i(m) → sublist(0, m, getarray(r0, i0, j0, st0)) = getarray(r0, i0, n→i(m), st0) ⊦ getarray(r, i, j, st) .first + sublist(0, m, getarray(r, i, j, st) .rest) = getarray(r, i, n→i(m) + 1, st)

Rule argument

[speclemmaarg: , vals ≠ [] → sublist(0, n + 1, vals) = vals .first + sublist(0, n, vals .rest) ,[vals, n] → [getarray(r, i, j, st), m] ,F ,list-del ,javavalues ,sub-zerorec ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < j → (¬ ¬ j ≤ i ↔ false)])], [EXTRATERMS : (getarray(r, i, j, st), m)]]